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### Functions of a Complex Variable

```Date: 10/05/2002 at 11:28:34
From: Jay
Subject: Functions of a complex variable

Show that w=f(z*), where z*=x-iy and f is a differentiable function,
is not an analytic function of z.
```

```
Date: 11/11/2002 at 11:43:14
From: Doctor Nitrogen
Subject: Re: Functions of a complex variable

Hi, Jay:

This gets pretty involved, so I hope you have time to read carefully.

The only proofs I could find somewhat related to my argument here can
be found on page 71, problem 2, and page 32, problem 5,in:

_Theory and Problems of Complex Variables_,
Murray Spiegel,
McGraw Hill, 1964, Schaum's Outlines, ISBN #07-060230-1.

Do you recall those delta(y)/delta(x) equations you had to use when
you first learned to calculate a derivative in calculus? That is, if
you recall, you had to compute a limit of something like

[f(x + delta(x)) - f(x)]/delta(x),

to find the derivative of something like

f(x) = x^2.

To prove that f(z*) fails to be analytic, you will have to do
something similar.

First of all, let

w = f(z*)

= ((u(x, y) + iv(x, y))*.

By the definition of derivative in Complex Variable Theory:

(d/dz*)f(z*)

= Limit         [f((z + delta(z))*) - f(z*)]/(delta(z*)).
delta(z*)->0

First, consider what happens when delta(y) = 0, lim delta(x)-> 0:

When (delta(y) = 0),

Limit         [f((z + delta(z))*) - f(z*)]/(delta(z*))
delta(x)->0

= Limit          {  [u(x + delta(x), y) - iv(x + delta(x), y)]
delta(x)->0

- [u(x, y) - iv(x, y)] } / delta(x)

= Limit          {   u(x + delta(x), y) - u(x, y)
delta(x)->0

- i[v(x + delta(x), y) + iv(x, y)] } /delta(x)

= Limit          { u(x + delta(x), y) - u(x, y) } / delta(x)
delta(x)->0

- [i(v(x + delta(x), y) - iv(x, y)] / delta(x)

[A]              = &u/&x - i&v/&x,

where "&this/&that" denotes "the partial derivative of this with
respect to that."

Now consider what happens when delta(x) = 0 and lim delta(y) -> 0:

When (delta(x) = 0),

Limit          [f(z + delta(z))*) - f(z*)]/(-i delta(y))
delta(y)->0

= Limit         {  [u(x, y + delta(y)) - iv(x, y + delta(y))]
delta(y_->0

- [u(x, y) - iv(x, y)] } / (-i delta(y))

= Limit         {  [u(x, y + delta(y)) - u(x, y)]
delta(y)->0

- [iv(x, y + delta(y) + iv(x, y)]} / (-i delta(y))

= Limit         {  [u(x, y + delta(y)) - u(x, y)]
delta(y)

+ [iv(x, y + delta(y) - iv(x, y)] } / (i delta(y))

[B]         = &u/&y + i&v/&y.

Now recall that for a complex valued function to be analytic, a
necessary and sufficient condition is for the Cauchy-Riemann equations
to be satisfied, that is:

[C]              &u/&x + i&v/&x = &v/&y - i&u/&y,

or
&u/&x = &v/&y,

&u/&y = -&v/&x.

But compare what you must have for f(z*) to be analytic from [A] and
[B]:

[A]             &u/&x - i&v/&x

[B]             &u/&y + i&v/&y,

or it must be that

[D]             &u/&x = &u/&y,

&v/&y = -&v/&x.

Clearly, [C] and [D] are not the same conditions, are they? But it is
known that [C], or the Cauchy-Riemann equations, must be satisfied,
so [C] is true, meaning the equations in [D] cannot be true, so f(z*)
can't be analytic.

You also might ask yourself, as another way to prove f(z*) fails to
be analytic, (I did not investigate this myself, but you might want
to) whether both

Re[f(z*)] and Im[f(z*)]

are harmonic functions, since every analytic function has two harmonic
functions associated with it, that is,

f(z) = u(x, y) + iv(x, y)

is an analytic function provided u(x, y) and v(x, y) are both
harmonic. If you can show Re[f(z*)] and Im[f(z*)] are not harmonic,
it would follow they are not analytic.

mathematics problem. You are welcome to return to The Math Forum and
Doctor Math whenever you have any math-related questions.

- Doctor Nitrogen, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
High School Calculus
High School Functions

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